Question

这道数学题不定积分怎么做

∫3x∧4＋2x∧2/x∧2＋1 dx

Answer 1

∫ 1/(2x2 - 1) dx = ∫ 1/[(√2x - 1)(√2x + 1)] dx = (1/2√2)∫ [(√2x + 1) - (√2x - 1)]/[(√2x - 1)(√2x + 1)] d(√2x) = (1/2√2)∫ [1/(√2x - 1) - 1/(√2x + 1)] d(√2x) = (1/2√2)(ln|√2x - 1| - ln|√2x + 1|) + C = ln|(√2x - 1)/(√2x + 1)|/(2√2) + C _______________________________ ∫ 1/(4x2 + 4x - 3) dx = ∫ 1/[(2x - 1)(2x + 3)] dx = (1/8)∫ [1/(2x - 1) - 1/(2x + 3)] d(2x) = (1/8)ln|(2x - 1)/(2x + 3)| + C ______________________________ ∫ (tan2x + tan?x) dx = ∫ tan2x(1 + tan2x) dx = ∫ tan2x ? sec2x dx = ∫ tan2x dtanx = (1/3)tan3x + C _________________________ ∫ cosx ? cos(x/2) dx = (1/2)∫ [cos(x + x/2) + cos(x - x/2)] dx = (1/2)∫ [cos(3x/2) + cos(x/2)] dx = (1/2)[(2/3)sin(3x/2) + 2sin(x/2)] + C = (1/3)sin(3x/2) + sin(x/2) + C ______________________________ ∫ (2x + 1)/(x2 - 2x + 2) dx = ∫ (2x - 2)/(x2 - 2x + 2) dx + 3∫ dx/(x2 - 2x + 2) = ∫ d(x2 - 2x + 2)/(x2 - 2x + 2) + 3∫ dx/[(x - 1)2 + 1] = ln|x2 - 2x + 2| + 3arctan(x - 1) + C _______________________________ ∫ x2/√(4 - x2) dx，x = 2sinθ，dx = 2cosθ dθ = ∫ 4sin2θ/(2cosθ) ? 2cosθ dθ = 2∫ (1 - cos2θ) dθ = 2θ - 2 ? 1/2sin2θ + C = 2arcsin(x/2) - 2(x/2)[√(4 - x2)/2] + C = 2arcsin(x/2) - (x/2)√(4 - x2) + C ______________________________ ∫ 1/[x2√(x2 - 1)] dx，x = secθ，dx = secθtanθ dθ = ∫ secθtanθ/(sec2θ ? tanθ) dθ = ∫ cosθ dθ = sinθ + C = √(x2 - 1)/x + C _______________________________ ∫ √(x2 - 1)/x dx，x = secθ，dx = secθtanθ dθ = ∫ tanθ/secθ ? secθtanθ dθ = ∫ tan2θ dθ = ∫ (sec2θ - 1) dθ = tanθ - θ + C = √(x2 - 1) - arcsecx + C = √(x2 - 1) - arccos(1/x) + C